Question

Write different oxidation states of manganese. Why is the +2 oxidation state of manganese more stable?

Answer 1

Hint: Manganese has 7 electrons in its valence shell. These electrons are present in the s and d subshell. Manganese can lose these electrons and show variable oxidation states.

Complete step by step answer:
> Manganese is a d-block element. It is transition metal. The atomic number of manganese is 25. Write the electronic configuration of manganese:
$$\left[\text{Ar}\right]3d^{5}4s^2$$.
> An atom of manganese can lose two to seven electrons to attain the oxidation states of +2, +3, +4, +5, +6 and +7. Thus the lowest oxidation state of manganese is +2 and the highest oxidation of manganese is +7. High oxidation states of manganese are usually seen in oxides. Thus, +7 oxidation state of manganese is shown in $$\text{M}{\text{n}}_{\text{2}}{\text{O}}_{\text{7}}$$.
> Potassium permanganate is an important oxidising agent. In potassium permanganate, the oxidation state of manganese is +7. During the redox reaction, the oxidation state of manganese decreases from +7 (in potassium permanganate) to +2 oxidation state. In this reaction, manganese itself is reduced and it oxidizes another substance.
> For +2 oxidation state, the electronic configuration is $$\left[\text{Ar}\right]3d^5$$. It has a half filled 3d subshell. Half filled (and completely filled) subshells are more stable than partially filled subshells. Hence, +2 oxidation state of manganese is more stable.

Note: Transition metals have incompletely filled inner d orbitals. Hence, they show variable valence. Half-filled and fully - filled subshells are more stable than any other electronic configurations.